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Consider $$S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}$$ and show that, for every positive integer $n$, $$S_n\ge\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\dfrac{\sqrt{2}}{2n}$$

I can prove a related upper bound: $$S_n=\sum_{k=1}^{n}\dfrac{1}{n\sqrt{1+\left(\dfrac{k}{n}\right)^2}}\le\int_{0}^{1}\dfrac{1}{\sqrt{1+x^2}}dx=\ln{(1+\sqrt{2})}$$ but I can't prove the lower bound.

Did
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math110
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3 Answers3

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The inequality to be proved is

$$\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}>\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\frac{\sqrt{2}}{2n}$$ equivalent to $$\frac{1}{n-1}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{1+(\frac{k}{n})^2}}>\ln{(1+\sqrt{2})}=\int_0^1\frac{1}{\sqrt{1+x^2}}.$$

LHS is the Riemann Sum of integral $\displaystyle\int_0^1\frac{1}{\sqrt{1+x^2}}$. When $n\to\infty$, LHS=RHS.

So what we have to prove is that LHS is monotonically decreasing with $n$.

Consider a funtion $f(x)=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+(1-x)^2}}$. Easy to know its 2nd derivative in $[0,1]$ is negative.

So we have $$f\Big(\frac{k}{n}\Big) > \Big(1-{k\over n}\Big)f\Big({k\over n+1}\Big)+{k\over n}f\Big({k+1\over n+1}\Big).$$ Sum over $n$, we get $$\sum_{k=1}^{n-1} f\Big(\frac{k}{n}\Big)>\sum_{k=1}^{n-1}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=1}^{n-1}\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^nf\Big(\frac{k}{n+1}\Big).$$ So $$\frac{1}{n-1}\sum_{k=1}^{n-1}\frac{1}{\sqrt{1+\Big(\dfrac{k}{n}\Big)^2}}=\frac{1}{2(n-1)}\sum_{k=1}^{n-1} f\Big(\frac{k}{n}\Big)$$ is monotonically decreasing with $n$. Q.E.D.

Hans
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Frank
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The bounds look like the first terms of an asymptotic series in 1/n. I'll develop such an asymptotic series. However, this is not a complete answer to the problem, as will be stated near the end.

First, use Gradshteyn 6.611, an integral involving a Bessel function, $ \int_0^\infty e^{-ax} J_0(bx) dx = 1/\sqrt{a^2+b^2}$ in the sum to get

$$S:=\sum_{k=1}^{n} 1/\sqrt{k^2+n^2} = \sum_{k=1}^{n} \int_0^\infty e^{-kx} J_0(nx) dx = \int_0^\infty J_0(nx) \dfrac{1-e^{-nx}}{(e^x-1)} dx.$$

Get a series in 1/n by the following, where $B_k$ are Bernoulli numbers,

$$S=\dfrac{1}{n} \int_0^\infty J_0(x)\dfrac{1-e^{-x}}{(e^{x/n}-1)} dx = \int_0^\infty dx \ J_0(x)\dfrac{1-e^{-x}}{x} \sum_{k=0}^\infty \Bigl(\dfrac{x}{n}\Bigr)^k \ \dfrac{B_k}{k!}. $$

Now do what an asymptoticist does, and interchange the sum and the integral. The first of the two resultant integrals are actually convergent:

$$S \sim \sum_{k=0}^\infty \Bigl(\dfrac{1}{n}\Bigr)^k \int_0^\infty J_0(x)(1-e^{-x})x^{k-1} \ dx = $$ $$\int_0^\infty J_0(x)\frac{1-e^{-x}}{x}dx - \frac{1}{2n} \int_0^\infty J_0(x)(1-e^{-x}) + \sum_{k=1}^\infty \dfrac{B_{2k}\ n^{-2k}}{(2k)!} \int_0^\infty dxJ_0(x)(1-e^{-x})x^{2k-1} $$

The first 2 integrals are $\log(1+\sqrt{2})$ and $1-1/\sqrt{2}$, respectively. The subsequent integrals are divergent, and a method of regularization is needed for them to make sense. I believe Chapter 4 of R. Wong, 'Asymptotic Approximation of Integrals' can be made to apply. In particular Lemma 4 of that chapter states that

$$\lim_{\epsilon\to0}\int_0^\infty J_\alpha(x)e^{-\epsilon \ x}x^{\mu-1}dx = \dfrac{\Gamma(\alpha/2 + \mu/2)2^{\mu-1}}{\Gamma(\alpha/2 - \mu/2 +1)} . $$

Thus the divergent part of the integral is set to zero by a property of the $\Gamma$ function. The portion with the exponential decay converges and can be expressed in terms of Gauss's hypergeometric function, and on collecting, we have the asymptotic series

$$S \sim \log(1+\sqrt{2}) -\frac{1-1/\sqrt{2}}{2n} - \sum_{k=1}^\infty \dfrac{B_{2k}\ n^{-2k}}{2k} F(k,k+1/2,1,-1) .$$

The standard process of summing to least term illustrates that this is a good approximation, even for small n. For example, including only the first asymptotic term in the infinite series gives you almost five digits precision for n as small as 2!

Now for purposes of satisfying the proposer's bound question, an explicit bound on the asymptotic series must be obtained. I have not done this, but the aforementioned chapter of Wong provides a way to do it. The first term past those given explicitly is $-1/(24\sqrt{2} \ n^2).$ The next term of $O(n^{-4})$ is also negative. The pattern appears to be $--++--++-- ...$ With a sufficiently precise bound on either of those first 2 negative terms, then one has proved that, for all sufficiently large n,

$$S > \log(1+\sqrt{2}) -\frac{1-1/\sqrt{2}}{2n} \approx 0.8814 - \dfrac{0.2929}{2n}.$$

The proposer's inequality can be written as $$S_p > \log(1+\sqrt{2}) -\frac{2\log(1+\sqrt{2})-\sqrt{2}}{2n} \approx 0.8814 - \dfrac{0.3485}{2n}.$$

$S_p$ asserts a more negative $O(1/n)$ term than in $S$, so it looks like some of the negativity of the $O(1/n^2)$ asymptotic term is being pushed into the $O(1/n)$ term. One can probably get to $S_p$ numerically by this approach, but the closed form for the coefficient suggests a different approach. (Unless one is being sneaky, found a numerical coefficient, and then found a closed form expression that approximates the coefficient.)

Did
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skbmoore
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  • So you memorized the Bessel function integral? Impressive. The approach is educational to me and looks promising. I will try to walk down this path. It would be great if you could carry out the rest of the proof. Thank you, skbmoore. – Hans Aug 14 '17 at 21:12
  • Check out Frank's correct and ingenious answer below using symmetrization, if you are still around. – Hans Sep 05 '17 at 19:32
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In a previous version of this answer, we first noted that the argument to get an upper bound of $$S_n=\frac1n\sum_{k=1}^nf\left(\frac{k}n\right)\qquad f(x)=\frac1{\sqrt{1+x^2}}$$ is simply that the function $f$ is decreasing on $x>0$ hence, for every $k$, $$\frac1nf\left(\frac{k}n\right)\leqslant\int_{(k-1)/n}^{k/n}f(x)dx$$ and, summing over $1\leqslant k\leqslant n$, one gets readily $$S_n\leqslant\sum_{k=1}^n\int_{(k-1)/n}^{k/n}f(x)dx=\int_0^1f(x)dx=\left.\arg\!\sinh x\,\right|_0^1=\arg\!\sinh1=\ln(1+\sqrt2) $$ Likewise, to get a lower bound, the most natural idea (in any case this is the idea we followed in the previous version of this answer...) might be to use once again the fact that the function $f$ is decreasing but this time, to the effect that, for every $k$, $$\frac1nf\left(\frac{k}n\right)\geqslant\int^{(k+1)/n}_{k/n}f(x)dx$$ hence, summing over $1\leqslant k\leqslant n-1$, one gets $$S_n-\frac1{\sqrt{2n^2}}\geqslant\sum_{k=1}^{n-1}\int^{(k+1)/n}_{k/n}f(x)dx=\int_{1/n}^1f(x)dx=\left.\arg\!\sinh x\,\right|_{1/n}^1=\arg\!\sinh1-\arg\!\sinh\left(\frac1n\right) $$ Unfortunately, $$\arg\!\sinh\left(\frac1n\right)>\frac1n\arg\!\sinh1$$ hence, to get the desired lower bound, this approach is doomed and another argument is required. (The trapezoidal rule might be all that is needed here, but I did not check.)

Nota: Thanks to user @Hans for having spotted this mistake (nearly 4 years later...).

Did
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  • It's nice.!Thank you +1 – math110 Oct 14 '13 at 17:20
  • But this answer misses the last term on the left, $\frac 1{\sqrt 2 n}$. What happens to that? – Hans Aug 01 '17 at 20:21
  • @Hans The OP asks to prove that $$\sum_{k=1}^na_k\geqslant b_n+\frac{\sqrt2}{2n}$$ for some numbers $b_n$ and $(a_k)$ such that $$a_n=\frac{\sqrt2}{2n}$$ and this answer proves that $$\sum_{k=1}^{n-1}a_k\geqslant b_n$$ It seems that "what happens" is that nothing is "missing", wouldn't you say? – Did Aug 01 '17 at 21:51
  • @Hans Did you downvote this by any chance? – Did Aug 04 '17 at 21:15
  • Oops, you are right. I missed $a_n$. No, I did not downvote. In fact, I upvoted before I doubled back for the "missing" $\frac1{\sqrt 2n}$, by then I could not cancel my upvote. I left the upvote there awaiting for your response. It is strange though I did not get a notified of your second last comment but only the last one just now. You should be able to see if there was any downvote, right? Was there? I could not think of any reason why if there was. – Hans Aug 04 '17 at 21:38
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    However, the direction of the first inequality sign should be reversed, so the answer is wrong. I am not going to downvote it until you correct it, unless you intend to leave it as it is. – Hans Aug 04 '17 at 22:37
  • @Hans Right, the approach seems fatally mistaken. I revised the post (which, unfortunately, cannot be deleted until the OP unaccepts it). Thanks for your careful reading. – Did Aug 05 '17 at 11:02
  • I tried the same failed approach as your latest one before I posted my last comment. It fails because the integral $\int_0^x$ is concave with respect to $x$. This makes the problem itself a more interesting one. You are welcome for the note of appreciation in your answer. – Hans Aug 07 '17 at 19:58
  • @Hans Downvote until corrected? Hm... seems Did shouldn't correct it in that case... – Simply Beautiful Art Aug 07 '17 at 21:41
  • @SimplyBeautifulArt: You meant to say "NOT downvote until corrected", right? Yes, it was not accurately written to convey my intention, despite the "unless" clause. I meant to say "I will leave the vote alone for now until further edit, unless..." – Hans Aug 07 '17 at 22:51
  • @Hans There appears to be new answers, if either of you care to look at them. – Simply Beautiful Art Aug 14 '17 at 20:11
  • @SimplyBeautifulArt: Simplifire's proof is wrong. See my comment below his answer. skbmoore's answer looks promising. – Hans Aug 14 '17 at 21:12
  • @SimplyBeautifulArt and Did: Check out the answer from Frank using with an ingenius symmetrization! Did could probably add another sentence to your note pointing to the correct answer from Frank. – Hans Sep 05 '17 at 19:31
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    @Hans I do appreciate your enthusiasm but please do not mess with my post for petty reasons ("Nota" is quite correct, you know). Likewise, Frank's answer is indeed very nice but why is it urgent that I add a note in my post saying so? – Did Sep 05 '17 at 19:44
  • My apology for my overreach in my excitement. I googled "nota" and found only that it was the plural form of the dorsal fin of a fish. So I thought "nota" was a typo in place of "note". Is "nota" Latin? – Hans Sep 05 '17 at 20:08
  • Regarding my suggestion for adding a sentence --- it is, of course, your prerogative to do whatever you see appropriate --- my rationale was the following. The OP had probably neglected his question page and would not change the acceptance check to the newly added correct answer. The sentence would clarify the situation for posterity. I should have stated clearly my intention for the friendly suggestion in my second last comment. – Hans Sep 05 '17 at 20:08